In determining automobile-mileage ratings, it was found that the mpg (X) for a certain model is normally distributed, with a mean of 33 mpg and a standard deviation of 1.7 mpg. Find the following:__________. a. P(X<30)
b. P(28c. P(X>35)
d. P(X>31)
e. the mileage rating that the upper 5% of cars achieve.
Answers
The upper 5% of cars have a mileage rating of 35.805 mpg
What is z score?
Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (raw score - mean) / standard deviation
Given; mean of 33 mpg and a standard deviation of 1.7
a) For < 30:
z = (30 - 33)/1.7 = -1.76
P(x < 30) = P(z < -1.76) = 1 - 0.8413 = 0.0392
b) For < 28:
z = (28 - 33)/1.7 = -2.94
P(x < 28) = P(z < -2.94) = 0.0016
c) For > 35:
z = (35 - 33)/1.7 = 1.18
P(x > 35) = P(z > 1.18) = 1 - P(z < 1.18) = 1 - 0.8810 = 0.119
d) For > 31:
z = (31 - 33)/1.7 = -1.18
P(x > 31) = P(z > -1.18) = 1 - P(z < -1.18) = 0.8810
e) The upper 5% of cars achieve have a z score of 1.65, hence:
1.65 = (x - 33)/1.7
x = 35.805 mpg
The upper 5% of cars have a mileage rating of 35.805 mpg
Find out more on z score at: brainly.com/question/25638875
Answer:
a) P(X < 30) = 0.0392.
b) P(28 < X < 32) = 0.2760
c) P(X > 35) = 0.1190
d) P(X > 31) = 0.8810
e) At least 35.7965 mpg
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
a. P(X<30)
This is the pvalue of Z when X = 30. So
has a pvalue of 0.0392.
Then
P(X < 30) = 0.0392.
b) P(28 < X < 32)
This is the pvalue of Z when X = 32 subtracted by the pvalue of Z when X = 28. So
X = 32
has a pvalue of 0.2776.
X = 28
has a pvalue of 0.0016.
0.2776 - 0.0016 = 0.2760.
So
P(28 < X < 32) = 0.2760
c) P(X>35)
This is 1 subtracted by the pvalue of Z when X = 35. So
has a pvalue of 0.8810.
1 - 0.8810 = 0.1190
So
P(X > 35) = 0.1190
d. P(X>31)
This is 1 subtracted by the pvalue of Z when X = 31. So
has a pvalue of 0.1190.
1 - 0.1190 = 0.8810
So
P(X > 31) = 0.8810
e. the mileage rating that the upper 5% of cars achieve.
At least the 95th percentile.
The 95th percentile is X when Z has a pvalue of 0.95. So it is X when Z = 1.645. Then
At least 35.7965 mpg
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Answers
Answer:
See the attached picture for detailed answer.
Step-by-step explanation:
See the attached picture for detailed answer.
Final answer:
The probability question from part (a) requires calculating the chance of getting all heads or all tails on multiple days in a year, which involves complex probability distributions. For part (b), using a Poisson distribution could be appropriate due to the rarity of the event and the high number of trials involved.
Explanation:
The question pertains to the field of probability theory and involves calculating the probability of specific outcomes when flipping a fair coin. For part (a), Jack flips a coin ten times each morning for a year, counting the days (X) when all flips are identical (all heads or all tails). The exact expression for P(X > 1), the probability of more than one such day, requires several steps. First, we find the probability of a single day having all heads or all tails, then use that to calculate the probability for multiple days within the year. For part (b), whether it is appropriate to approximate X by a Poisson distribution depends on the rarity of the event in question and the number of trials. A Poisson distribution is typically used for rare events over many trials, which may apply here.
For part (a), the probability on any given day is the sum of the probabilities of all heads or all tails: 2*(0.5^10). Over a year (365 days), we need to calculate the probability distribution for this outcome occurring on multiple days. To find P(X > 1), we would need to use the binomial distribution and subtract the probability of the event not occurring at all (P(X=0)) and occurring exactly once (P(X=1)) from 1. However, this calculation can become quite complex due to the large number of trials.
For part (b), given the low probability of the event (all heads or all tails) and the high number of trials (365), a Poisson distribution may be an appropriate approximation. The mean (λ) for the Poisson distribution would be the expected number of times the event occurs in a year. Since the probability of all heads or all tails is low, it can be considered a rare event, and the Poisson distribution is often used for modeling such scenarios.
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Answers
cos x = 0.899
Answers
Answer:
the answer is 8a
Step-by-step explanation:
becuase the numbers are simply being added, you can just add 3+5.
for example, say a=3
3(a)+5(a)=3(3)+5(3)
9+15=24
but 3+5(a)=8(a)
8(3)=24
Answer:
8a2
Step-by-step explanation:
Answers
Answer:
120 degrees.
Step-by-step explanation:
If it is a regular 6 sided figure then each interior angle is 180 - (360/6)
= 120 DEGREES.
Answers
The probability that it will take more than 6 minutes for all the customers in line to check out is 0.40.
We are given the probability distribution of x, the number of customers in line at a supermarket express checkout counter.
Moreover, we are given that each customer takes 3 minutes to check out.
It means that if there are 0 customers in line, i.e., x=0, then it will take 0 minutes for all the customers currently in line to check out.
If there is 1 customer in line, i.e., x=1, then it will take 3 minutes for all the customers currently in line to check out.
If there are 2 customers in line, i.e., x=2, then it will take 6 minutes for all the customers currently in line to check out.
If there are 3 customers in line, i.e., x=3, then it will take 9 minutes for all the customers currently in line to check out.
If there are 4 customers in line, i.e., x=4, then it will take 12 minutes for all the customers currently in line to check out.
If there are 5 customers in line, i.e., x=5, then it will take 15 minutes for all the customers currently in line to check out.
From above we note that if there are 3 or more customers in the line, then it will take more than 6 minutes (note that the case of check out time equal to 6 minutes is not included when we want 'more than 6 minutes') for all the customers currently in line to check out.
Thus, required probability is given by:
P(more than 6 minutes for all the customers currently in line to check out) = P(x ≥ 3)
= P(x=3) + P(x=4) + P(x=5)
= 0.20 + 0.15 + 0.05
= 0.40
Therefore, the probability that it will take more than 6 minutes for all the customers in line to check out is 0.40.
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Final answer:
Without specific information on the total number of customers or the distribution of customers in line, we cannot calculate a specific probability for it to take more than 6 minutes for all customers to check out, given that each customer takes 3 minutes.
Explanation:
The question is about the probability that it will take more than 6 minutes for all the customers in line to check out, given that each customer takes 3 minutes. The time it takes for all the customers to check out is determined by the number of customers in line. If there are two or more customers in line, it will definitely take more than 6 minutes for all of them to check out, because the checkout time is 3 minutes per customer.
So, the question of probability relates to the likelihood of there being two or more customers in line. Without information on the total number of customers, or the distribution of customers in line, we cannot calculate a specific probability.
Please note, this is a practical application of topics in probability and queue theory, involving concepts like mean arrival rate and service rate.
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$____
Answers
86-15+24=95$
Have a great day hope it helps